I can't figure out how the Quadrature (Q) component is populated when BIST PRBS is enabled on AD9361.

This post describes the BIST PRBS algorithm:

Exact HDL implementation of AD9361 PRBS

Each successive value is this: dout = {din[14:0], ((^din[15:4]) ^ (^din[2:1]))};

I've confirmed that this works for the In-Phase (I) component, but it appears something else is done for Q.

At any sample, I noticed that bits 7:0 for Q are the bit-reverse of bits 7:0 for I. Maybe that's just coincidence due to the PRBS.

If someone can explain how to compute/check Q for the BIST PRBS, that'd be great. Thanks

I think I figured this out.

Q uses the same polynomial, but with bits in reverse order (including the bits for past polynomial state).

I: [past 4][ bits 11:0 ]

Q: [ bits 11:0 ][ past 4]

The new bit (computed via XOR) is shifted in at the MSB, pushing data out the RHS in little endian notation.